# Applications of Projectile Motion Projectile Motion in

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```					                                        Applications of Projectile Motion - 5.5
1. Projectile Motion in a Vertical Direction:
Let v 0 be the initial velocity of an object which travels in a vertical direction at the initial height h 0 . Let h!t"
represent the height of the object at the time t. Then by Newton’s second law of motion: F ! ma, where F is
the sum of the forces acting on an object, m is the mass of the object and a is the acceleration of the object, we
have ignoring air resistance
a!t" ! h "" !t" ! !g the gravity, g ! 9. 8 meters/second, g ! 32 feet/second
v!t" ! h " !t" ! !gt # v 0
h!t" ! ! 1 gt 2 # v 0 t # h 0
2
Example Velocity of a diver at impact: If a diving board is h 0 feet above the water surface and a diver starts
with initial velocity v 0 feet/second, without considering air resistance, what is the diver’s velocity
at impact?
The diver’s velocity at impact is the velocity when the diver hits the water surface. First we find the time when
the diver hits the water surface: h!t" ! 0
! 1 gt 2 # v 0 t # h 0 ! 0, ! 1 g t 2 ! 2 v 0 t ! 2h 0 ! 0
g        g
2                          2
!v 0 \$ v 2 ! 4 ! 1 g !h 0 "
0       2                !v 0 \$ v 2 # 2gh 0
0
t!                                 !          !g
2 !1g
2

v 0 # v 2 # 2gh 0
0
t0 !            g           seconds v 2 # 2gh 0 " v 0
0

v!t 0 " ! !g!t 0 " # v 0 gives the diver’s velocity at impact. Example, let h 0 ! 120 feet, v 0 ! 0.
2!32"!120"
t0 !                   ! 2. 738 613 seconds
32
v!2. 738 613" ! !!32"!2. 738 613" ! !87. 635 62 ! !87. 635 62 3600 ! !59. 75 mph
5280

Example Suppose that a raindrop falls from a cloud 3000 feet above the ground. How fast would be
raindrop be falling when it hits the ground if we ignore the air resistance? What is the velocity of
the raindrop at that time? Is it reasonable?
y!t" ! ! 1 !32"t 2 # 3000. y!t" ! !8!2t 2 ! 375" ! 0, t !
2
375
2
! 13. 693 06 seconds
y " !t" ! !32t, y " !13. 69" ! !32!13. 69" ! ! 438. 08 feet/second
! !438. 08 3600 ! ! 298. 69 mph - it is not possible.
5280

2. Projectile Motion in Vertical and Horizontal Directions:
Let v 0 be the initial velocity at an angle ! of an object which travels in both horizontal and vertical directions.
Let y!t" represent the vertical position and x!t" represent the horizontal position. Then by Newton’s second law
of motion: F ! ma, where F is the sum of the forces acting on an object, m is the mass of the object and a is the
acceleration of the object, we have ignoring air resistance
y "" !t" ! !g, the gravity, g ! 9. 8 meters/second, g ! 32 feet/second, x "" !t" ! 0
y " !t" ! !gt # v 0 sin !, x " !t" ! v 0 cos !
y!t" ! ! 1 gt 2 # v 0 sin ! t # y 0 , x!t" ! v 0 cos ! t # x 0
2
How height and how far can the object go? When v "y !t" ! 0, the object reaches its highest position.

1
y " !t" ! 0 # t 0 ! v 0 sin !,
g
v 2 sin 2 !
y max ! y!t 0 " ! ! 1 g v 0 sin ! # v 0 sin ! v 0 sin ! # y 0 ! 1 0 g
2
g                      g                                  # y0
2                                                 2
When t ! 2t 0 , the object stops horizontally. So,
2v 2 sin ! cos !
x max ! x!2t 0 " ! v 0 cos ! 2 v 0 sin ! # x 0 !
g
0
g          # x0

Example An object is launched at angle ! ! " from the horizontal with initial speed v 0 ! 98 meters/second.
6
Determine the landing time and the horizontal range of the projectile.
y!t" ! ! 1 !9. 8"t 2 # 98 sin " t # 0 ! !4. 9t 2 # 49t ! !4. 9t!t ! 10" ! 0, t ! 0, t ! 10
2                    6
t ! 0 seconds is the launching time and t ! 10 seconds is the landing time.

120

100

80

60

40

20

0
2     4   t   6       8       10

y!t"
"
x!10" ! 98 cos     6
t # 0 ! 49 3 !10" ! 848. 704 9 meters.

Example Suppose that a tennis player hits a serve from a height of 8 feet at an initial speed of 120 mph and
at an angle of 5 # below the horizontal. The serve is in if the ball clears a 3’-high net that is 39’
away and his the ground in front of the service line 60’ away. Determine whether the serve is in or
out.
1 mile ! 880 % 2 % 3 ! 5280 feet, 120 mph ! 120 5280 ! 176. 0
3600
y!t" ! ! 1 !32"t 2 # 176. 0 sin!!5 # "t # 8, x!t" ! 176. 0 cos!!5 # " t
2
Find how long it takes for the ball horizontally to get 39’ away:
x!t" ! 176. 0 cos!5 # "t ! 39, t !         39        ! 0. 222 437 4 seconds
176. 0 cos!5 # "
Find the height of the ball at that time:
y!0. 222 437 4" ! ! 1 !32"!0. 222 437 4" 2 # 176. 0 sin!!5 # "!0. 222 437 4" # 8
2
! 3. 792 167 feet & 39
So the ball is out.

2

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